# In how many ways can 5 prizes be given away to 4 boys?

• RChristenk
RChristenk
Homework Statement
In how many ways can ##5## prizes be given away to ##4## boys when each boy is eligible for all the prizes?
Relevant Equations
Concepts in permutations
I first thought obviously the solution would be ##5^4##. This made intuitive sense to me because this solution means the first boy can choose one out of five prizes, the second boy can choose one out of five (even if it's the same prize as the first boy's since each boy is eligible for all prizes), and so on until the fourth boy. So in essence the exponent ##4## is the number of boys.

But the correct answer is the opposite: ##4^5##. This also makes sense. The first prize can be given to one of the four boys, the second prize can be given to one of the four boys (even if it's the first boy again since each boy is eligible for all prizes) and so on until the fifth prize. In this case the exponent would be the number of prizes.

Since both answers make intuitive sense, how can they be numerically different when both are dealing with four boys and five prizes? Where did I go wrong? Thanks.

Hi,
With 4^5 > 5^4 there are a lot of possibilities you missed. Can you find one ?

##\ ##

You first approach only gives 4 prizes away, with each student getting one prize. Your second approach gives all 5 prizes away, with students allowed to get multiple prizes to the extent that all 5 prizes can go to one student.
From the statement of the problem that you give, it is not possible to know exactly what the correct answer is.

The fact that each boy is eligible for all prizes means that the prizes are interchangeable, so that the problem reduces to "stars and bars": how may ways can you order 5 stars and 4 bars?

I understand now. The answer is ##4^5## because "when each boy is eligible for all the prizes" means it is possible one boy gets all five prizes and the remaining three boys get nothing. So the first prize can be given to one out of four boys. The second prize can be given to one out of four boys (including the one that received the first prize) and so on.

##5^4## would mean each boy got a prize, and each prize can be given out repetitively, ie. there are five different kinds of prizes but there is an infinite supply of each one. Hence the first boy gets to receive one out of the five prizes, the second boy gets to receive one out of the five prizes (including the same prize as the first boy) and so on.

So the first case is there are only five prizes, and it is possible for one boy to take them all and the remaining three left with nothing. The second case is five different prizes with infinite supply each, every single boy gets one prize, and the prize can be the same as another boy's.

How many ways can two prizes be given to one boy?

You can see it as each of the 4 boys having 5 choices each. This is too, the number of functions from a 5-element set into a 4-element set, which is what such an assignment is.

RChristenk said:
I understand now. The answer is ##4^5## because "when each boy is eligible for all the prizes" means it is possible one boy gets all five prizes and the remaining three boys get nothing.
That is the correct interpretation of the problem, assuming that the problem was stated so carefully and read carefully. Unfortunately, you can not always be sure that the exact wording can be taken literally. IMO, it would be better if the problem statement stated clearly that each boy could have any number of prizes. Combinatorial problems should not require such careful reading when a little more description could make it clear.

phinds and WWGD
pasmith said:
The fact that each boy is eligible for all prizes means that the prizes are interchangeable, so that the problem reduces to "stars and bars": how may ways can you order 5 stars and 4 bars?
I could be mistaken, but 4 slots means 3 bars instead of 4.

Edit: The prizes are apparently not identical, so the method of 5 stars and 3 bars would be incorrect. Looks like @FactChecker has it right.

Last edited:
FactChecker
FactChecker said:
That is the correct interpretation of the problem, assuming that the problem was stated so carefully and read carefully. Unfortunately, you can not always be sure that the exact wording can be taken literally. IMO, it would be better if the problem statement stated clearly that each boy could have any number of prizes. Combinatorial problems should not require such careful reading when a little more description could make it clear.
I remember a question in a Qual exam along the lines of " What can you say about the Baire Category theorem in the Real line? Great to have such an open-ended question in an exam that will shape your life.

FactChecker
Edit: The prizes are apparently not identical, so the method of 5 stars and 3 bars would be incorrect. Looks like @FactChecker has it right.
I hadn't noticed that. That is one more thing that the statement of the problem leaves for us to guess.

I have some sympathy for the question setter in this case. The phrase "eligible for each prize" suggests something like a prize for each of five academic subjects or sporting events. I find it difficult to interpret that as five identical prizes. Although of course, that is a possible interpretation.

So, if boy 1 gets prize A first then later prize B, this is counted as different than getting B first then A? I would have counted final states.

Paul Colby said:
So, if boy 1 gets prize A first then later prize B, this is counted as different than getting B first then A? I would have counted final states.
Those two must be the same.

Let's sort this out for 3 prizes (A,B, C) and two boys. Each solution is defined by what the first boy gets.

A,B and C (1 solution)
Any two prizes (3 solutions)
Any one prize (3 solution)
No prizes (1 solution)

That's a total of 8 solutions, which is ##2^3##.

Note that we can map each solution to a three digit binary number, where each digit represents a specific prize, and 0 and 1 ( or and 2) represents the boy who won that prize.

This can be done in general to give ##b^p## solutions, where ##b## is the number of boys etc.

Paul Colby said:
So, if boy 1 gets prize A first then later prize B, this is counted as different than getting B first then A? I would have counted final states.
Presumably the (distinct) prizes are given given out in a predetermined order, and we don't need to add extra statistics to count the 5 factorial ways that the prizes could be ordered.

Paul Colby
Presumably the (distinct) prizes are given given out in a predetermined order, and we don't need to add extra statistics to count the 5 factorial ways that the prizes could be ordered.
Or, track the prize orderings and remove them at the end which is what I did to get deconfused.

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